Math 10

Now a structure seen in figure (ii) will be formed, which is a cylinder.

Now, length of AB or DC = Circumference of the circular base of the cylinder.

= 2πr

Breadth (BC) = Height of the cylinder (h)

Area of rectangular sheet ABCD = Curved surface area of cylinder

Now,

Curved surface area of the cylinder = Area of rectangular sheet.

= AB × BC = 2πrh

∴ Curved surface area of cylinder (CSA) = 2πrh

Total surface area of the cylinder

(TSA) = 2 × area of base + curved surface area

= 2πr2 + 2πrh = 2πr(r+h)

Volume of the cylinder

As we know that the volume of a cuboid

= Area of base × height,

Then, volume of cylinder (V) = πr2×h = πr2h

Hence, V = A × h

Volume of materials contained by a hollow cylinder

If R be the external radius, r the internal radius and 'h' the height of a hollow cylinder,
then

External volume of the cylinder (V1) = πR2h rR
Internal volume of the cylinder (V2) = πr2h

∴ The volume of material contained h

by the hollow cylinder = V1 – V2

= πR2h – πr2h

= πh (R2 – r2)

Surface area and Volume of half cylinder.

Lets take a cylinder having radius 'r' and height 'h'. Lets

divide the cylinder into two equal halves by splitting it

vertically. h

Lets observe the half part of the cylinder. Its both end
contains semi circle. Its one side contains a rectangular
surface having length 'h' and breadth '2r' or 'd'. Another
part is curved which is half of curved surface area of cylinder.

Hence, total surface area of half cylinder is the sum of area of two semi circles, a rectangle

Oasis School Mathematics-10 95

and half curved part of cylinder.

So, TSA = 2 × semi circle + area of rectangular surface + area of curved surface

= 2 × 1 πr2 + 2rh + πrh.
2

Hence, TSA of half cylinder = πr2 + 2rh + πrh.

Again, volume of half cylinder is half of the volume of cylinder.

So, volume = 1 πr2h.
2

Remember !

• Volume of a cylinder = Area of base × height

• Volume of a cylinder = πr²h

• Curved surface area of a cylinder = Perimeter of the base × height

• Curved surface area of a cylinder = 2πrh

• Total surface area of a cylinder = CSA + 2 × area of base

• Total surface area of a cylinder = 2πr(r + h)

• CSA of half cylinder = πrh.

• TSA of half cylinder = πrh + 2rh + πr2

• Volume of half cylinder = 1 πr2h.
2
1
Note : 1 litre = 1000 cm3, 1 litre = 1000 m3.

Worked Out Examples


Example: 1

A cylinder has radius 7 cm and height 10 cm. Find its
(i) curved surface area (ii) total surface area (iii) volume
Solution: Here,

Radius of the cylinder (r) = 7 cm

height (h) = 10 cm
We have,

(i) Curved surface area of the cylinder (CSA) = 2πrh

= 2 × 22 × 7 × 10 cm²
7

= 440 cm²

(i) Total surface area of the cylinder (TSA) = 2πr (r + h)

= 2 × 22 × 7 (7 + 10) cm2
7

= (44 × 17) cm2

= 748 cm2

96 Oasis School Mathematics-10

(ii) Volume (V) = πr2h

= 22 × 72 × 10 cm3
7

= 1540 cm3

Example: 2

The diameter of a cylindrical log of wood is 28 cm. If the area of its curved surface is
1188 cm2, find its height.

Solution:

Here, Diameter of the cylindrical log (d) = 28 cm

Radius of the cylindrical log (r) = d = 14 cm
2

Curved surface area (CSA) = 1188 cm2

or, 2πrh = 1180 cm²
or,
or, 2 × 22 × 14 × h = 1188 cm²
7

616 h = 1188 × 7

∴ h = 1188 × 7 cm
616

= 13.5 cm
∴ Required height of the cylindrical log =
13.5 cm.

Example: 3
The capacity of a cylindrical can is 1.54 litre. If its radius is 7cm, find its height.
Solution: Here,

Volume of cylinder (V) = 1.54 litre

= 1.54 × 1000cm3 = 1540 cm3

Radius (r) = 7 cm

We have, V = πr2h

1540 = 22 × 72 × h
7

1540 = 154 × h

h = 1540 = 10 cm
154

∴ Height of the cylindrical can is 10 cm.

Example: 4

The external and internal radii of a hollow cylindrical vessel 21 cm high are 7.77 cm
and 7.35 cm. respectively. Find the volume of material contained by the vessel.

Solution: Here,
External radius of cylindrical vessel (R) = 7.77 cm.

Oasis School Mathematics-10 97

Internal radius of the vessel (r) = 7.35 cm
Height (h) = 21 cm.

∴ Volume of material (V) = π h (R2 – r2)

= 22 × 21 [(7.77)2 – (7.35)2] cm3
7

= 66 × 6.3504 cm3

= 419.12 cm3

Example 5 :

Find the curved surface area and total surface area of given half cylinder.

Solution:

In the given half cylinder 21 cm

Radius (R) = 21 cm = 10.5 cm.
2

We have, CSA of half cylinder = πrh. 50 cm

= 22 × 10.5 × 50 cm2
7

= 1650 cm2

Again, TSA of half cylinder = πr2 + πrh + 2rh

= 22 (10.5)2 + 22 (10.5)2.50 + 2 × 10 × 5 × 50
7 7

= 346.50 + 1650 + 1050

= 3045.60 cm2
Example: 6

The sum of the radius and height of a cylinder is 21 cm and its curved surface area is
616 cm2. Find its (i) height, (ii) radius, (iii) TSA.

Solution:

Given, r + h = 21cm or r = 21–h ..........(i)

CSA = 616 cm2

We have,

CSA of a cylinder = 2πrh

616 = 2π(21 – h).h
or,
or, 616 = 2 × 22 h (21 – h)
7

616 × 7 = 44h (21 – h)

or, 616×7 = 44 × 21h – 44h2

or, 44 × 14 × 7 = 44 (21r – h2)

or, 14 × 7 = 21h – h2

or, h2 – 21h + 98 = 0

or, h2 – (14 + 7)h + 98 = 0

or, h2 – 14h – 7h + 98 = 0

98 Oasis School Mathematics-10

or, h(h – 14) – 7 ( h – 14) = 0

or, (h - 14) (h – 7) = 0

Either, h-14 = 0, then, h = 14 cm

or, h-7 = 0, then, h = 7 cm

When h = 14cm,

From (i), r = (21–14)cm = 7cm

Now,

TSA of cylinder = 2πr(r + h)

=2× 22 × 7 × 21 = 924 cm2
7

When h = 7 cm

from (i), r = (21 – 7) cm = 14 cm

Now, TSA of cylinder = 2πr (r + h)

=2× 22 × 14 × 21
7

= 44 × 2 × 21 = 1848 cm2.
Example: 7

The curved surface area and volume of a cylinder are 264 cm2 and 924 cm3 respectively.
Find the radius, height and TSA of the cylinder.

Solution:
Given, CSA of the cylinder = 264 cm2

2πrh = 264 cm2

h = 264 ................(i)
2πr

Again, volume = πr2h

924 = πr2h

or, h = 924 ................(ii)
πr2

From (i) and (ii)

226π4r = 924
πr2
924
or, 132 = r

or, 132r = 924 h= 264 = 6cm
44
924
∴ r = 132 = 7cm We have,

From equation (i) TSA of cylinder = 2πr (r+h)

h = 264 =2× 22 × 7 (7+6)
2πr 7

= 264 =44 × 13 cm2

2 × 22 × 7 =572 cm2
7

Oasis School Mathematics-10 99

Exercise 6.1

1. (a) Write the formula to find the area of the base of a cylinder.

(b) Write the formula to find the perimeter of the base of a cylinder.

(c) If the perimeter of the base and height of the cylinder are given, write the formula
to find the curved surface area of the cylinder.

(d) Write the formula to find the curved surface area of a cylinder in terms of 'r'
and 'h'.

(e) If the CSA and area of the base of a cylinder are given, write the formula to find the
total surface area of the cylinder.

(f) Write the formula to find the total surface area of a cylinder in terms of 'r' and 'h'.

(g) If the area of base and height of the cylinder are given, write the formula to find
the volume of the cylinder.

(h) Write the formula for the volume of a cylinder in terms of 'r' and 'h'.

2. Find the curved surface area, total surface area and volume of the following solids.

(a) 7cm (b) 28cm (c) 21cm

20cm 35cm 20cm



3. (a) Find the curved surface area of a cylinder whose perimeter of the base is 60 cm and
the height 30 cm.

(b) Find the curved surface area of a cylinder whose circumference of the base is 44 cm
and the height 20 cm.

(c) Find the total surface area of a cylinder whose circumference of the base is 88 cm
and the sum of the radius and the height is 40 cm.

(d) If the sum of the radius and height of the cylinder is 10 cm and the circumference
of the base is 308 cm, find the total surface area of that cylinder.

(e) The circumference of the base of a cylindrical drum is 44 cm. If the sum of its
radius and the height is 17cm, find its total surface area.

(f) The area of a base of the cylinder is 154 cm² and the height is 20 cm. Find its
volume.

(g) Find the volume of a cylinder whose area of the base is 616 cm² and the height is
30 cm.

4. (a) The volume of a cylindrical solid is 1,320 m³. If the area of the base is 264 m²,
calculate the height of the solid.

100 Oasis School Mathematics-10

(b) The area of the base of a cylinder is 154 cm2 and its volume is 9240 cm3. Find the
height and radius of the base.

(c) A cylindrical water tank contains 46,200 liters of water. If the radius of the base is
3.5 m , find its height.

(d) Volume of a cylinder is 7,700 cm3. If its height is 50 cm, find its radius.
(e) Volume of cylinder is 24,640 cm3. If its height is 40 cm, find its curved surface area.

5. (a) 50 circular plates, each of a radius 7 cm and thickness 5 mm are placed one above
the other to form a cylindrical shape. Find the volume of the cylinder so formed.

(b) 35 coins of Re. 1 are kept one above the other. If the diameter of each coin is 1cm
and thickness is 2mm, find the volume of the object so formed.

6. (a) The circumference of the base of a cylinder is 88 cm and height 40 cm. Find its total
surface area.

(b) The circumference of the base of a cylindrical object is 22 cm and height is 40 cm.
Find the total surface area of the object.

(c) The perimeter of a circular base of a cylinder is 132 cm. If the height of the cylinder
is 18 cm, find its total surface area and volume.

(d) The diameter of a cylindrical log of wood is 28 cm. If the area of the curved surface
is 1188 cm², find its volume.

(e) The sum of the height and the radius of the base of a cylinder is 34 cm. If the total
surface area of the cylinder is 2,992 cm², find the radius of the base.

7. Find the volume of given half cylinder.

(a) (b) (c)

10 cm 20 cm 28 cm

7 cm 14 cm 50 cm

8. Find CSA and TSA of given half cylinder. (c) 7 cm

(a) (b) 4 0 cm

14 cm 20 cm
28 cm

30 cm

9. (a) If the height and radius of a cylindrical piece of wood are equal, and the curved
surface area is 308 cm2, find its volume.

(b) The curved surface area of a cylinder whose height is equal to the radius of the
base 1,232 sq. cm. Find the total surface area of the cylinder.

(c) The radius and height of a cylinder are in the ratio 5:7. If the volume of the cylinder
is 550 cm3, find the radius of the base of the cylinder.

Oasis School Mathematics-10 101

(d) The ratio of the height and radius of the base of a right cylinder is 4:3. If its curved
surface area is 528 cm², find its volume.

10. (a) Curved surface area of a cylinder is 1,760cm2. If the sum of the radius and height
is 34 cm, find its

(i) height (ii) radius (iii) TSA

(b) The sum of the diameter and height of a cylinder is 21cm. If its CSA is 308 cm2, find its

(i) height (ii) radius (iii) TSA (iv) volume.

11. (a) The curved surface area of a solid cylinder is equal to 2 of the total surface area
3
of the same cylinder. If the total surface area is 924 cm2, find its volume.

(b) The curved surface area of the cylinder is 2 of its total surface area. If its total
3

surface area is 1,212 cm², find its radius and height.

12. (a) The external and internal diameter of a metallic hollow cylinder of length 25 cm
are 8 cm and 6 cm respectively. Find the volume of the metal used in it.

(b) A cylindrical pipe 2.4 cm long has thickness of 0.5 cm and an external diameter of
12 cm. Find the volume of the pipe.

13. (a) The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3.
Find the diameter and height of the pillar.

(b) The curved surface area of a cylinder is 308 cm2 and its volume is 1,078 cm3, find
its height and radius.

Answers

1. Consult your teacher 2. (a) 880 cm2, 1,188 cm2, 3,080 cm2 (b) 3,080cm2, 4,312cm2, 21,560cm3
(c) 1,320 cm2, 2,013 cm2, 6,930 cm3 3. (a) 1,800cm2 (b) 880cm2 (c) 3,520cm2 (d) 3,080cm2
(e) 748cm2 (f) 3,080cm3 (g) 18,480cm3 4. (a) 5m (b) 60cm, 7cm (c) 1.2m (d) 7cm
(e) 3,520cm2 5. (a) 3,850cm3 (b) 5.5cm3 6. (a) 4,752cm2 (b) 957cm2 (c) 5,148cm2,
24,948cm3 (d) 8,316cm3 (e) 14cm 7. (a) 192.5 cm3 (b) 1540 cm3 (c) 15400 cm3

8. (a) 660 cm2, 1234 cm2 (b) 1760 cm2, 3496 cm2 (c) 220 cm2, 398.50 cm2
9. (a) 1,078cm3 (b) 2,464cm2 (c) 5cm (d) 2,095.43cm3
10. (a) (i) 20cm (ii) 14cm (iii) 2,992cm2 or, (i) 14cm (ii) 20cm (iii) 4,274.29cm2 (b) (i) 7cm

(ii) 7m (iii) 616cm2 (iv) 1,078cm3 11. (a) 2,156cm3 (b) 8.01cm, 16.04cm
12. (a) 550 cm3 (b) 43.37cm3 13. (a) 14m, 6m (b) 7 cm, 7cm

6.3 Sphere and Hemisphere

Let's collect objects like a football, tennis ball, globe, marble and observe their surface
and shape.



102 Oasis School Mathematics-10

• The surface of each object is curved.
• These are spherical objects.

A'

BO A OA = radius = OA'
O is the center of the sphere.

A sphere is formed by rotating the circle about the diameter.

Surface area of the sphere



d

dd

Take a cylinder whose diameter is equal to the height. Take a sphere whose diameter is
equal to the diameter of the cylinder.

Wrap the curved surface of the cylinder with a rope. With the same rope, the surface of
the sphere can be wrapped.

Then,

Surface area of sphere = Curved surface area of cylinder

= 2πrh

= 2πrd [ h = d]

= 2πr (2r) = 4πr2

∴ Surface area of sphere = 4πr2

Since the entire surface of the sphere is curved, curved surface area and total surface
area of sphere are equal.

Volume of a sphere

Take a hollow cylinder whose diameter is equal to its height. Take a sphere whose
diameter is equal to the diameter of the cylinder.

Fill the cylinder with water. Now, immerse the sphere completely into the cylinder.

dd

d dd

Collect the displaced water from the cylinder. You can see that the volume of displaced
2
water is 3 of the volume of water in the cylinder.

Oasis School Mathematics-10 103

Then, 2
3
Volume of sphere = volume of cylinder

= 2 πr2h
3

= 2 πr2 (2r)
3

= 4 πr3
3

∴ Volume of sphere = 4 πr3
3

Great circle and hemisphere

When a sphere is divided into two equal halves, each half is called a r

hemisphere. It has a circular face whose radius is equal to the radius of

the sphere. The circular face is called the great circle.

Curved surface area of a hemisphere: = 1 of surface area of sphere
2

= 1 × 4 πr2 = 2πr2
2

Total surface area of a hemisphere = Curved surface area+Area of circular face

Total surface area of a hemisphere = 2πr2 + πr2 = 3πr2

Volume = 1 of volume of sphere
2

= 1 × 4 πr3 = 2 πr3
2 3 3

Volume of materials contained by hollow spherical shells

Let R be the external radius and r the internal radius of a hollow spherical shell.

Then, the external volume of the shell (V1) = 4 πR3
3

the internal volume of the shell (V2) = 4 πr3
∴ 3

The volume of the material contained by the hollow spherical shell

= external volume of the shell–internal volume of the shell

= 4 πR3 – 4 πr3
3 3
4
= 3 π (R3 – r3)

Remember !

• Volume of a sphere = 4 πr3 • CSA of a sphere = 4πr²
3
2
• TSA of a sphere = 4πr² • Volume of a hemisphere = 3 πr3

• CSA of a hemisphere = 2πr² • TSA of a hemisphere = 3πr²

104 Oasis School Mathematics-10

Worked Out Examples

Example: 1

Find the total surface area and the volume of a sphere whose diameter is 28 cm.
Solution:

Given, Diameter of the sphere (d) = 28 cm

Radius (r) = 28 = 14 cm
We have, 2

Total surface area of the sphere (TSA) = 4πr²

= 4 × 22 × 14 × 14 cm2
7

= 2464 cm²

Again, we have, Volume (V) = 4 πr3
3

= 34 × 22 × (14)3 cm3
7

= 43 × 22 × 14×14×14 cm3
7

= 11,498.66 cm3
Example: 2

Find the curved surface area, total surface area and volume of a hemisphere whose
radius is 7 cm.

Solution:

We have, Given, radius (r) = 7 cm

Curved surface area (CSA) = 2πr²

= 2 × 22 × (7)2
7
= 308 cm²

Total surface area (TSA) = 3πr²

= 3 × 22 × (7)²
7

= 462 cm²

Volume (V) = 2 πr3
3

= 2 × 22 × (7)3
3 7

= 718.67 cm3

Example: 3

The circumference of the great circle of a sphere is 44 cm. Find its surface area.
Solution:

Circumference of great circle (C) = 44 cm.

Oasis School Mathematics-10 105

or, 2 πr = 44 cm.
or,
∴ 2× 22 × r = 44 cm.
7
r = 7 cm

Since radius of the sphere = radius of great circle

Surface area of sphere = 4 πr2

= 4 × 22 × 72
7
= 616 cm2.

Example: 4
If the total surface area of the sphere is 22,176 cm², find
(i) its radius
(ii) its volume
Solution:

Given, total surface area (TSA) = 22,176 cm²

Radius (r) = ?

We have,

Total surface area (TSA) = 4πr²

or, 22176 = 4 × 22 r²
7

or, 7 × 22176 = 4 × 22 r²

or, r² = 22176×7
4×22

or, r² = 1764

or, r² = (42)²

or, r = 42 cm

Again, we have, Volume (r) = 4 πr3
3

= 34 × 22 (42)3
7

= 3,10,464 cm3

Example: 5

Three spherical iron balls of radii 1 cm, 6 cm and 8 cm are melted down and recast
into a single solid sphere. Find the radius of the single sphere so formed.

Solution:
Let r1, r2 and r3 be the radii of three given spheres.

Here, r1 = 1 cm, r2= 6 cm, r3 = 8 cm

106 Oasis School Mathematics-10

We have, volume of the first sphere (V1) = 4 πr13
3

= 4 π(1)3 cm3
3

Volume of the second sphere (V2) = 4 πr23
3

= 4 π(6)3 cm3
3
4
Volume of the third sphere (V3) = 3 πr33

= 4 π(8)3 cm3
3

Let 'r' and V be the radius and volume of the new sphere so formed.

We have, V = V1 + V2 + V3

4 πr3 = 4 π(1)3 + 4 π(6)3 + 4 π(8)3
3 3 3 3

or, 4 πr3 = 4 π [(1)3 + (6)3 + (8)3]
3 3
or,
or, r3 = 729
∴
r3 = (9)3

r = 9 cm

Hence, radius of the new sphere is 9 cm.

Exercise 6.2

1. (a) Write the formula to calculate the great circle of a sphere.
(b) Write the formula to calculate the surface area of a sphere.
(c) If 'r' be the radius of a sphere, what is the volume of the sphere?
(d) Write the formula for the volume of sphere in terms of its diameter.
(e) If 'A' be the area of the great circle of a sphere, what is the surface area of the sphere

in terms of A?
(f) Write the formula of the surface area of a sphere in terms of 'd'.
2. (a) Write the formula to calculate the curved surface area of a hemisphere.
(b) Write the formula to calculate the total surface area of a hemisphere.
(c) If 'A' be the area of the great circle of a hemisphere then what is the CSA and TSA

of the hemisphere.

(d) Write the formula of the volume of a hemisphere in terms of 'r' and in terms of 'd'.

3. Find the curved surface area, total surface area and volume of the following solids.

(a) (b) (c)

3.5cm 14cm 15cm





Oasis School Mathematics-10 107

4. Find the curved surface area, total surface area and volume of the following solids.

(a) (b) (c) 14cm
7cm 7cm



5. (a) If the volume of a sphere is 9 cm3 , find its radius.
16 π2
9π
(b) If the volume of the spherical object is 2 cm³, find its diameter.

(c) If the volume of a sphere is 38,808 cm3, find its curved surface area.

(d) Find the total surface area of a sphere whose volume is 1372 π cm³.
3
(e) The volume of a sphere is 12,348 π cm3, find its surface area.

6. (a) Surface area of a sphere is 1,764 π cm². Find its radius.

(b) If the total surface area of a sphere is 616 cm², find its volume.

(c) Surface area of a sphere is 4 π cm². Find its volume.

(d) If the circumference of the great circle of a sphere is 440 cm, find its surface area.

7. (a) A spherical ball of lead is melted and made into smaller balls of half the radius of
the original radius. How many such balls can be made?

(b) Three metallic spheres of radii 2 cm, 12 cm and 16 cm respectively are melted and
reformed to a single sphere. Find the radius of new sphere so formed.

(c) Three hemispheres of radii 3cm, 4cm and 5cm are melted to form a hemisphere.

Find the radius of the hemisphere so formed.

8. (a) If the volume of a hemisphere is 16 π cm³ , find its radius.
3

(b) If the volume of a hemisphere is 19,404 cm³, find its total surface area.

(c) If the curved surface area of a hemisphere is 308 cm², find its total surface area.

(d) The perimeter of the plane circular of a hemisphere is 176 cm. Calculate the total
surface area and the volume of the hemisphere.

(e) Find the total surface area and the volume of a hemisphere having a curved surface
area 2,772 cm2.

9. (a) A solid metallic sphere of radius 21 cm is cut into two halves. Find the total surface

area of the two hemispheres so formed.

(b) The total surface area of a solid sphere is 616 cm². Two hemispheres are formed

when it is cut into two equal parts. Find the total surface area of each hemisphere.

10. (a) The external and internal radii of a hollow spherical metallic shell are 35 cm. and
14 cm. respectively. Calculate the volume of the metal contained by the shell.

(b) External diameter of a hollow sphere is 14 cm. If the thickness of the metal contained
by it is 2 cm, find the volume of the metal contained by it.

108 Oasis School Mathematics-10

11. (a) 27 solid iron spheres, each of radius 'x' cm, are melted to form a sphere with radius
'y' cm. Find the ratio x : y.

(b) 8 solid iron spheres with diameter 'a' cm each are melted to form a sphere with
diameter 'b' cm. Find the ratio a : b.

12. (a) If a metallic sphere having volume 45π cm3 is melted to form a cylinder of height
5 cm, then find the radius of that cylinder.

(b) If a metallic hemisphere having volume 1,540 cm3 is melted to form a cylinder
having height 10 cm, find the radius of this cylinder.

(c) If a metallic cylinder having volume 18π cm3 is melted to form a hemisphere find
the radius of the hemisphere.

Answers

1. Consult your teacher 2. Consult your teacher

3. (a) 154 cm2, 154 cm2, 179.67 cm3 (b) 616 cm2, 616 cm2, 1437.33 cm3

(c) 2828.57 cm2, 2828.57 cm2, 14142.85 cm3

4. (a) 308cm2, 462cm2, 718.67cm3 (b) 77cm2, 115.5cm2, 89.83cm3 (c) 1232cm2, 1848cm2, 5749.33 cm3

5. (a) 3 cm (b) 3 cm (c) 5544 cm2 (d) 6,16 cm2 (e) 5,544 cm2
4π
4π
6. (a) 21cm (b) 1437.33cm3 (c) 3 cm3 (d) 61,600cm2 7. (a) 8 (b) 18 cm (c) 6cm

8. (a) 2cm (b) 4158cm2 (c) 462cm2 (d) 7,392cm2, 45,994.67cm3 (e) 4,158cm2, 19,404cm3

9. (a) 8,316cm2 (b) 462cm2 10. (a) 1,68,168cm3 (b) 913.52 cm3 11. (a) 1: 3 (b) 1:2

12. (a) 3 cm (b) 3 cm (c) 3 cm

6.4 Cone

Look at the given pictures.

In our day-to-day life, we are familiar with some objects like
an ice-cream container, birthday cap, funnel to pour kerosene,
etc. These geometrical figures have unique similarity as they
have a circular base tapering to a point. Such figures are called
a cone.

These objects are conical in a shape. Hence they are a cone. O
The given figure represents a cone, where 'O' is the vertex.
slant height (l)
OA and OB are slants heights of the cone. It is denoted by 'l'. height (h)

OC is its height. It is denoted by 'h'.

BC is its radius. It is denoted by 'r' AC B

Here, radius(r)
∆OCB is a right-angled triangle. Hence, OB2 = OC2 + BC2

i.e., l2 = h2 + r2

Oasis School Mathematics-10 109

Activity

Draw a circle and draw a sector AOB O

as shown in the figure. O

OAB

Cut the sector AOB with the

help of scissors. B

A

Fold this sector such that OA just

coincides with OB. The structure so formed is cone.



Surface area of cone
Take a hollow cone and cut it along its slant height as shown in the figure.

l"

123 4 5 6 7 8
r 2πr

[Circumference of a cone is equal to the arc length of the sector]

You will get a sector as shown in the figure. The arc length of the sector is equal to 2πr
as it is equal to the circumference of the base of the original cone. Its radius is equal to l.

πr

1357 l
246 8

πr

Now, length of this shape = 2πr = πr
2
breadth = l

Area of this shape = πr × l

= πrl

Which represents the CSA of the cone.

∴ Curved surface area of the cone (CSA) = πrl.

Again, Total surface area = Curved surface area + Area of base

= πrl + πr2

= πr (l + r)

Hence, TSA of a cone = πr (r + l).

110 Oasis School Mathematics-10

Volume of cone

Take a hellow cone and a hollow cylindrical jar having
the same radius and height.

Fill the cone with water and pour it into the cylindrical

jar. You can see that three cones full of water just fill
the cylinder.

Then,

Volume of cylinder = 3 × Volume of cone

Volume of cone = 1 Volume of cylinder
3

= 1 (πr2h) = 1 πr2h
3 3

Hence, volume of cone = 1 πr2h
3

= 1 (area of base) × height
3

= 1 A×h
3

Remember !

• CSA of cone = πrl

• TSA of cone = πr (r + l)

• Volume of cone (V) = 1 πr²h
3
1
• Volume of cone (V) = 3 area of base × height

• TSA of cone = CSA + Area of base

• Relation among r, l and h is l2 = h2 + r2


Worked Out Examples

Example: 1

The area of the circular base of a cone is 100 cm2 and its height is 18 cm. Find the
volume of the cone.

Solution: Here,
Area of the circular base of the cone (A) = 100 cm2

height of the cone (h) = 18 cm

So, Volume of the cone (v) = 1 area of circular base × height
3

= 1 × 100 × 18 cm3 = 600 cm3
Example: 2 3

The radius of the circular base of a cone is 7 cm and slant height is 20 cm. Find the
curved surface area and total surface area.

Oasis School Mathematics-10 111

Solution: Here,

Radius of the circular base (r) = 7 cm

Slant height (l) = 20 cm

We have, the curved surface area (CSA) = π r l.

= 22 × 7 × 20 cm2
7

= 440 cm2

Again, and total surface area (TSA) = π r (l + r)

= 22 × 7 (20 + 7)
7

= 22 × 27 cm2 = 594 cm2

Example: 3

Find the volume of the given cone. P
h
Solution:

Here, Radius of circular base (r) = 8 cm 10 cm

Slant height (l) = 10 cm

so, height of cone (h) = l2 – r2 = 100 – 64 O
8 cm
= 36 = 6 cm

We have, volume (V) = 1 π r2 h
3

= 1 × 22 × 82 × 6 cm3 = 402.28 cm3
Example: 4 3 7

1430 1980
Curved surface area and total surface area of a cone are 7 cm² and 7 cm²
respectively. Find the slant height of the cone.

Solution:
1430

Given, curved surface area = 7 cm2
1980

Total surface area = 7 cm2

We have, TSA = CSA + area of base

1980 1430
or, area of base = 7 – 7

or, 550
πr² = 7

or, 22 r² = 550
7 7

or, 550
r² = 22

or, r2 = 25 or, r = 5 cm



112 Oasis School Mathematics-10

Again, we have,

CSA = πrl

1430 22 × 5 × l
or, 7 = 7
or,
22 × 5 l = 1430
1430
l =
22 × 5

or, l = 13 cm

Slant height = 13 cm

Example: 5

Find the total surface area of the given cone, whose semi-vertical angle is 30° and the
radius is 7 cm.

Solution: A C 7 cm B
30°
In the given cone
Radius (r) = 7 cm O

∠BOC = 30°

In right-angled triangle BOC, BC

sin 30° = OB
r
1 = l
2

or, l = 2r

or, l = 2×7 cm

= 14 cm

We have, TSA of a cone = πr (r + l)

= 22 × 7 (7 + 14)
7

= 22 × 21 cm² = 462 cm²

Exercise 6.3

1. (a) If 'A' be the area of the base of the cone and 'h' its height, write the formula to
calculate the volume of the cone.

(b) Write the formula of the volume of a cone in terms of 'r' and 'h'.

(c) Write the relation among radius (r), height (h) and slant height (l) of a cone.

(d) Write the formula for the curved surface area of a cone.

(e) Write the formula for the total surface area of a cone in terms of CSA of cone and
area of base.

(f) What is the formula of the TSA of a cone in terms of 'r' and 'l'?

Oasis School Mathematics-10 113

2. Find the curved surface area and total surface area of a cone having,
(a) radius of the base = 7 cm, slant height = 25 cm .
(b) radius of the base = 10 cm, slant height = 15 cm.
3. Find the volume of the cone having
(a) radius of the base = 7 cm, height = 24 cm.
(b) radius of the base = 5 cm, height = 12 cm.

(c) area of the base = 300 cm², height = 10 cm.

4. (a) If the slant height and radius of a cone are 10 cm and 6 cm respectively, find its
curved surface area.

(b) The diameter of the circular base of a cone is 20 cm and its slant height 21 cm. Find
its curved surface area and total surface area.

(c) The circumference of the base of a cone is 22 cm, and the sum of its radius and
slant height is 16 cm. Find its total surface area.

(d) If the circumference of the base is 44 cm and the slant height is 15 cm, find the
total surface area of the cone.

5. (a) The radius of the circular base of a cone is 14 cm and its height is 10 cm. Find its
volume.

(b) If the area of the base of the cone is 250 cm² and the height is 5 cm, find its volume.

(c) If the perimeter of the base of the cone is 88 cm and the vertical height is 10 cm,
find the volume of the cone.

6. Find the curved surface area and total surface area of the following solids.

(a) (b) (c) (d)

6 cm25 cm12 cm
15 cm

7 cm 12 cm 5 cm 16 cm
B
7. Find the volume of the given solids. B (d)

(a) (b) (c)

12 cm 25 cm

5 cm
24 cm

25 cm
5 cm O A
O 3 cm A 14 cm

8. (a) Acone whose radius of base is 7 cm, has a volume of 1,232 cm³. Find the height of the cone.
(b) If the volume of the given cone is 23,100 cm³ and its height is 50 cm, what is the

radius of the base?
(c) A cone 21 cm high has a volume of 1,078 cm3. Calculate the radius of its base.

114 Oasis School Mathematics-10

(d) The volume of the cone is 324 π cm³ and the radius of the base is 9 cm. Find the

curved surface area of the cone. P

9. (a) The total surface area of the cone whose radius is 10 cm is 594
cm2. Find the slant height of the cone.

(b) The curved surface area of a cone is 8,880 cm². If the slant of O A
the height is 100 cm, find the height of the cone.

(c) The total surface area and the diameter of the base of the cone are 300π cm² and
24 cm respectively. Find the slant height of the cone.

(d) Total surface area of the cone is 2,816 cm². If the radius is 14 cm, find

(i) slant height of the cone (ii) height of the cone (iii) volume of the cone

(e) Curved surface area of a cone is 550 cm². If its diameter is 14 cm, find its volume.

10. (a) In the given cone, its radius and height are equal. If the volume of the cone is 9702
cm³, find the height of the cone.

(b) The radius and the height of the cone are in the ratio 3:4. If the volume of the cone
is 96π cm3, find the slant height.

(c) The ratio of height and slant height of a cone is 24:25. If its volume is 1,232 cm3, find
its lateral surface area and total surface area.

(d) The vertical height of a cone is 3 times its radius. If its volume is 27π cm3, find its
radius, height and slant height.

11. (a) The total surface area and the curved surface area of a cone are 704 cm² and 550
cm² respectively. Find the slant height of the cone.

(b) The total surface area and the curved surface area of a cone are 2,816 sq. cm and
2,200 sq. cm respectively. Find the radius and slant height of the cone.

12. (a) If the given figure of the solid cone has a base diameter of 10 cm 13 cm
and slanting height 13 cm, find:

(i) area of the curved surface.

(ii) total surface area.

(iii) volume of the cone. [π = 22 ] 10 cm
7

(b) If the given figure of the solid cone has a base radius of 7 cm and 25 cm
slanting height 25 cm, find:

(i) area of the curved surface.

(ii) total surface area. 7 cm

(iii) volume of the cone. [π = 22 ]
7

(c) In the given figure, the semi-vertical angle of the cone is 30°. 30°
If its diameter is 14 cm, find 14 cm
(i) its CSA. (ii) its TSA. (iii) its volume.

Oasis School Mathematics-10 115

(d) In the given figure, the semi-vertical angle of the cone is 30°. 20cm 30°
If its height is 20 cm, find

(i) its CSA. (ii) its TSA. (iii) its volume.

14 cm

13. (a) If the water filled in the cylinder is poured 24 cm
into the cone, to what height will the surface
of the water reach?

14 cm

(b) A cylinder of height 40 cm and diameter 14 cm is melted to form a right circular
cone of height 30 cm. Find the radius of the cone.

Answers
1. Consult your teacher 2. (a) 550 cm2, 704 cm2 (b) 471.42 cm2 785.71 cm2
3. (a) 1,232 cm3 (b) 314.28 cm3 (c) 1,000 cm3 4. (a) 188.57 cm2 (b) 660 cm2, 974.28 cm2
(c) 176 cm2 (d) 484 cm2 5. (a) 2053.33 cm3 (b) 416.67 cm3 (c) 2,053.33cm3
6. (a) 330 cm2 484cm2 (b) 471.43 cm2, 584.57 cm2 (c) 204.29 cm2, 282.86 cm2 (d) 251.42 cm2,

452.57 cm2 7. (a) 314.28cm3 (b) 1,232 cm3 (c) 37.71cm3 (d) 1,232cm3 8.  (a) 24cm
(b) 21cm (c) 7cm (d) 424.29cm2 9. (a) 8.9 cm (b) 96cm (c) 13 cm (d) (i) 50 cm (ii) 48 cm
(iii) 9,856cm3 (e) 1,232 cm3 10. (a) 21cm (b) 10cm (c) 550cm2, 704cm2, (d) r = 3cm, h=9 cm,
l = 9.49cm 11. (a) 25cm (b) 14 cm, 50cm 12. (a) (i) 204.28 cm2 (ii) 282.86 cm2
(iii) 314.28 cm3 (b) (i) 550 cm2 (ii) 704 cm2 (iii) 1,232 cm2 (c) (i) 308 cm2 (ii) 462cm2
(iii) 622.38 cm3 (d) (i) 761.9cm2 (ii) 1,180.94cm2 (iii) 2,793.65cm3 13. (a) 4 cm (b) 14cm

6.5 Combined Solids


Here we study the volume, curved surface area and total surface area of a combined solid

formed by a cylinder and a cone, cylinder and a hemisphere, and a cone and a hemisphere.

Combined solid of cylinder and cone:

This is the combined solid formed by a cylinder and a cone.


V olume of this combined solid = Volume of cylinder + Volume of cone

C urved surface area = CSA of cylinder + CSA of cone
Total surface area = Area of base + CSA of cylinder + CSA of cone

Combined solid of cone and hemisphere:

This is the combined solid of a cone and a hemisphere.
Volume of this combined solid = Volume of cone + Volume of hemisphere
CSA of this combined solid = CSA of cone + CSA of hemisphere
TSA of this combined solid = CSA of cone + CSA of hemisphere

116 Oasis School Mathematics-10

C ombined solid of cylinder and hemisphere

This is the combined solid of a cylinder and a hemisphere.



Volume of given combined solid = Volume of cylinder + Volume of hemisphere
CSA of given combined solid = CSA of cylinder + CSA of hemisphere
TSA of given combined solid = Area of base + CSA of cylinder + CSA of hemisphere

Note:

Volume of Combined solid = Sum of the volumes of solids
CSA of a combined solid = Sum of the areas of visible curved parts
TSA of combined solid = Sum of the areas of all visible parts

Worked Out Examples

Example: 1 14 cm 25 cm
Find the volume of the given combined solid. 50 cm
Solution:

This is the combined solid of a cylinder and a cone.

For cylinder, Diameter (d) = 14 cm

Radius (r) = 14 cm = 7cm
2

Height (h) = 50 cm

We have, volume of cylinder (V1) = π r2 h

= 22 ×7² × 50 = 7,700 cm3
7

For Cone, Radius (r) = 14 = 7 cm
2

Slant height (l) = 25 cm

We have,

h = ² – r²
(25)² – (7)²
= 625 – 49

= 576 = 24 cm

= 1 π r2 h
We have, 3

Volume of cone (V2) =

Oasis School Mathematics-10 117

= 1 × 22 (7)² × 24 = 1,232 cm3
3 7

Now,

Volume of given combined solid = V1 + V2

= 7700 cm3 + 1,232 cm3 = 8,932 cm3
Example: 2

Find the curved surface area and total surface area of the given solid.

Solution: 50 cm

This is the combined figure of a cylinder and a hemisphere.

For the hemisphere, 64 cm

Radius (r) = 64 cm – 50 cm = 14 cm

We have,

CSA of hemisphere = 2 π r2

= 2 × 22 × (14)²
7

= 1,232 cm²

For the cylinder, radius (r) = 14 cm

Height (h) = 50 cm

We have, CSA of cylinder = 2πrh

= 2 × 22 × 14 × 50
7
N ow,
CSA of given solid = 4,400 cm²

= CSA of hemisphere + CSA of cylinder

= 1,232 cm² + 4,400 cm²

= 5,632 cm²

Again, area of base (circle) = π r2

= 22 × 14²
7

= 616 cm²

Now, TSA of given solid = CSA of hemisphere + CSA of cylinder + area of circle

= 1,232 cm² + 4,400 cm² + 616 cm² = 6,248 cm²

Example: 3

A tent in the form of a right circular cylinder of height 5 m is 10 m
surmounted by a cone of slant height 10 m. If the diameter of the
base of the tent is 14 m, find the total surface area of the tent. 5m
14 m
Solution:

Total surface area of the tent is the sum of the CSA of the cone and
CSA of the cylinder in this figure.

118 Oasis School Mathematics-10

For the cylinder,

Radius (r) = 14 = 7 m
2

Height (h) = 5 m

We have, = 2 π rh
CSA of cylinder
= 2× 22 × 7 × 5 = 220 m²
7
For Cone, Radius (r)
= 7 m

Slant height (l) = 10 m

We have, CSA of cone = πrl


= 22 × 7 × 10 = 220 m²
7

Now, TSA of given tent = CSA of cylinder + CSA of cone

= 220 m² + 220 m² = 440 m²

Example: 4 A
25 cm
The total surface area of the given combined solid is 858 cm². If the
slant height of the cone is 25 cm, find the length of their common
radius.

Solution:

Given figure is the combined figure of a cone and a hemisphere.

Here,

CSA of hemisphere = 2 π r²

CSA of cone = π rl

= πr × 25 = 25 πr cm²

Now, TSA of given figure = CSA of hemisphere + CSA of cone

or, 858 = 2 πr² + 25 πr
or,
or, 858 = 2 × 22 × r² + 25 × 272× r
7

858 × 7 = 2 × 22r² + 25 × 22r

or, 39 × 7 = 2r² + 25r

or, 2r² + 25r – 39 × 7 = 0

or, 2r² + (39 – 14)r – 39 × 7 = 0

Oasis School Mathematics-10 119

or, 2r² + 39r – 14r – 39 × 7 = 0

or, r(2r + 39) – 7 (2r + 39) = 0

or, (2r + 39) (r – 7) = 0
Either,
or, 2r + 39 = 0 ⇒ r = – 39 (not possible)
2
r – 7 = 0 ⇒ r = 7 cm

∴ Common radius = 7 cm

Exercise 6.4

1. Find the volume of the given combined solids.

(a) (b) (c) 30 cm
13 cm
4 cm 25 cm
10 cm
3 cm 10 cm

140 cm 54 cm

A A 14 cm (f) 10 cm A

(d) (e)

r = 7 cm P O r = 8 cm B

O 31 cm

25 cm A'

A' (i) 54 cm

A (h)
(g)
r= 10 cm

P 9 cm 12 cm P' 14 cm

O

(j) A' 37 cm 15 cm 40 cm

14 cm

2. Find the curved surface area (CSA) and total surface area (TSA) of the given solids.

(a) (b) (c)
140 cm

12 cm 10 cm 10 cm 13 cm 10cm

80 cm 24 cm 152 cm
57 cm
(d) 50 cm 25 cm (e) 63 cm (f)
14 cm
74 cm 140 cm

120 Oasis School Mathematics-10

(g) (h) 14 cm

37 cm 31 cm
30 cm

22 m 33m

3. (a) Find the total surface area of the given tent.

120m

(b) A tent is of the shape of a right circular cylinder of height 4 cm and radius 14 cm
with a cone of height 6 cm. Find the total surface area and volume of the tent.

(c) Height of a cylinder is 3 times its radius. It has hemispheres on both ends. If the
radius of the object is 7cm, find its volume.

(d) Find the volume of a combined solid containing a cylinder and hemisphere, where
the height of the cylinder is 50 cm and the height of whole solid is 64 cm.

4. (a) The base area of the cylinder is 100 cm2 and the height of the cylinder 3 cm
is 3 cm. If the volume of the whole solid is 500 cm3, find the height of
the solid so formed.



(b) The base of the cylinder is 125 cm² and the height of the

cylinder is 3 cm. If the volume of the whole

solid is 625 cm³, find the height of the solid. 3 cm

36 cm

(c) The volume of the given solid object is 38,808 cm3. 240 cm
Find the volume of the cone.



x cm

(d) If the volume of the given combined solid is 792 cm³, 6 cm
find the value of x.

Oasis School Mathematics-10 121

(e) The given figure is combined figure of a cone and a hem- 42 cm
isphere of diameter 42 cm. The volume of the solid is

38,808 cm3. Find the height of the cone.

5. (a) Curved surface area of the given solid is 924 cm². Find its height 14 cm
and total surface area. 10 cm




(b) Curved surface area of a given combined solid is

880 cm². Find the radius of the solid.

15 cm

6. (a) If the curved surface area of the given combined solid is
858 cm², find its common radius.

25 cm

(b) If the total surface area of the given combined solid 25 cm 30 cm
is 2,024 cm², find its common radius.

(c) The curved surface area of the given combined solid is
836 cm². Find its common radius and total surface area.

12 cm

Answers
1. (a) 138.27 cm3 (b) 11,314.29 cm3 (c) 5,852 cm3 (d) 1,950.67 cm3 (e) 1,950.67 cm3

(f) 1,475.05 cm3 (g) 2,200 cm3 (h) 770 cm3 (i) 30,389.33 cm3 (j) 5,338.67 cm3
2. (a) 3,205.71 cm2, 3,318.85 cm2 (b) 958.58 cm², 1,037.15 cm² (c) 4,604.28 cm², 4,682.86 cm²
(d) 2750 cm², 2904 cm² (e) 55440 cm², 67914 cm² (f) 2,508cm2, 2,662cm² (g) 985.6 cm², 985.6

cm² (h) 858 cm², 858 cm² 3. (a) 19,799.99 (b) 1,022.18cm2,, 3,696cm3 (c) 4,671.33cm3

(d) 36,549.33cm3 4. (a) 9 cm (b) 9 cm (c) 1,848 cm3 (d) 26 cm (e) 42 cm
5. (a) 21 cm, 1,078 cm² (b) 7 cm 6. (a) 7 cm (b) 7 cm (c) 7 cm, 990 cm²

122 Oasis School Mathematics-10

Unit

7 Prism and Pyramid

7.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.
Let's compare the similarities and dissimilarities of the given figures.
• What is the shape of each figure on the base?
• What type of surface do they have? Curved or flat?

Fig (i) Fig (ii)

Is it possible to find their surface area and volume using a similar formula?

7.2 Triangular Prism

The given figure is a triangular prism. It has a triangular base. Cross section A'
The base of this prism is ∆ABC or ∆A'B'C, which are two
Base

congruent triangles. A C'
It has three rectangular surfaces. B'

Its rectangular surfaces are ABB'A', BCC'B' and ACC'A'. B C

Surface Area of Triangular Prism A

The given figure is a triangular prism. A

∆ABC is its base. Three sides of ∆ABC c b

are 'a', 'b' and 'c'. Let 'h' be the height B a C AcB a Cb A
of the prism. If we unfold figure (i), we
A' h h h hh
will get a structure like figure (ii).

The surface area of this prism is the

sum of area of 2 triangular surfaces B' C' A' B' C' A'
A' Fig (ii)
and the area of 3 rectangular surfaces. Fig (i)

The area of 3 rectangular surfaces is its

lateral surface area.

Oasis School Mathematics-10 123

From figure (ii),

Area of rectangle ABB'A' = h × c

Area of rectangle BCC'B' = h × a

Area of rectangle CAA'C' = h × b

∴ Area of all rectangular surfaces = h × c + h × a + h × b

= h(a+b+c)

= perimeter of base × height

Hence, lateral surface area (LSA) = Area of all rectangular surfaces

= p×h

Total surface area (TSA) = LSA + 2 × area of ∆ABC

= LSA + 2 × area of base.

Remember !

LSA = p × h
TSA = LSA + 2 × area of base
Area of cross section = Area of base

Volume of Triangular Prism

Given figure is a cuboid having three dimensions l, b and h.

If we cut this cuboid along its diagonal, we get two right triangular prisms.

Area of base of given cuboid = l × b = A

Area of base of a right-triangular prism = 1 (l × b) = 1 A = A1
2 2

Volume of the cuboid = (l × b × h)

Volume of the right triangular prism = 1 (l × b) × h l bh
2

= 1 (l × b) × h.
2

= A1 × h.
∴ Volume of a prism = Area of base × height.

Remember !

In a triangular prism,
• Base area = Area of triangle = Area of cross section
• Lateral surface area = Perimeter of base × Height of the prism
• Total surface area = 2 Base area + Lateral surface area = 2∆ + p × h
• Volume = Area of triangular base × Height of prism

124 Oasis School Mathematics-10

Note:

Given figure is a square based prism.

• Base area = a2

• Lateral Surface Area (LSA) = Perimeter of base × height h

= 4ah a
a
• TSA = LSA + 2 × area of base = 4ah + 2a2
• Volume = Area of base × height = a2h

Worked Out Examples

Example: 1

Find the volume of a prism whose area of base is 20 cm2 and height is 10 cm.
Solution:
Here, Area of base of the prism (A) = 20 cm2

height (h) = 10 cm.

∴ volume of prism (v) = A × h = 20 cm2 × 10 cm = 200 cm3.

Example: 2

Find the lateral surface area and total surface area of the given prism.

Solution: D
A

4 cm E

B 3 cm C 8 cm

The triangular base of the given prism is a right-angled triangle.

AC = AB² + BC² = 4² + 3² = 25 = 5 cm

∴ Perimeter of triangular base (P) = (4 + 3 + 5) cm = 12 cm

Height of the prism (h) = 8 cm

We have, Lateral surface area = P × h = 12 × 8 = 96 cm2

Area of triangle ABC = 1 AB × BC
∴ ∆ = 2
1
2 ×4×3 = 6 cm2

We have, total surface area of the prism (TSA) = 2∆ + p.h

= (2× 6 + 12 × 8) cm2

= (12 + 96) cm2
= 108 cm2.

Oasis School Mathematics-10 125

Example: 3

Find LSA, TSA and volume of the given triangular prism.

Solution: 4 cm

10 3 .cam²
4
The triangular base of the given prism is an equilateral triangle.

Area of the equilateral triangle ABC = 3 .(as²ide)2 = 3 .×a²16 cm2 = 4 3 .cam² 2
4 4 4

Perimeter of the triangle (P) = (4 + 4 + 4) = 12 cm.

Height of prism (h) = 10 3 .cam²
∴ Lateral surface area (LSA) = P× h4

= 12 × 10 3 .cam² 2 = 120 3 .cam² 2.
And total surface area (TSA) = 2 ∆ + P. h4 4


= (2 o×f4ba4s3e.+×a1²h2e0ig4h3t).acm² 2 = 128 3 .cam² 2
area 4
Again, we have Volume (V) =

= 4 3 .×a1² 0 3 .cam² 3 = 120 cm3.
44

Exercise 7.2

1. (a) If 'A' be the area of the base and 'h' the height of the prism, write the formula to
calculate the volume of the prism.

(b) Write the formula to calculate the lateral surface area of the prism.
(c) Write the formula to calculate the total surface area of the prism.
(d) If 'a' be the length of a side of the base of a prism, which is an equilateral triangle,

and h the height of the prism, write the formula to calculate the LSA of the prism.
2. (a) Find the volume of a prism whose area of base is 75 cm² and height is 10 cm.
(b) A prism has a base area 50 cm2 and volume 600 cm3. Find its height.
(c) If the perimeter of the base of a triangular prism is 40 cm and the height is 25 cm,

find its lateral surface area.
(d) If the lateral surface area of a triangular prism is 540 cm² and the perimeter of the

base is 30 cm, find the height of the prism.
(e) The area of the base of a triangular prism is 120 cm² and the lateral surface area is

200 cm². Find the total surface area of the prism.

(f) The lateral surface area and total surface area of a triangular prism are 240 cm²
and 600 cm², find the area of the base.

126 Oasis School Mathematics-10

(g) The area of the base of a triangular prism is 60 cm². If its total surface area is
400 cm², find its lateral surface area.

3. Find the volume of the given triangular prisms.

(a) (b) A' (c) A'

6 cm A A

8 cm B' C' B' C'
9 cm 10 3 c.am² 12 3 c.am²
12 cm B 4 cm C B 8 cm C 4
4

(d) (e) 30 cm D (f) 50 cm
10 2cm
7 cm 7 cm A
6 cm
10 cm
EGF

B 12 cm C

4. Find the lateral surface area and total surface area of the given triangular prisms.

(a) (b) A A' (c)

10 cm 4 cm 3 cm
12 cm
B'
4 cm9 cm B 5 cm C 30 cm C' 4 cm
3 cm
(d) (e) (f)

4.5 cm 10 cm 6 cm 50 cm
13 cm

12 cm

7 cm

8 cm

5. (a) If a prism is 10 cm high with its triangular base having sides 3 cm, 4 cm and 5 cm,
find its total surface area.

(b) A triangular prism has length 20 cm and its triangular base is an isosceles
right-angled triangle having equal sides 6 cm. Find its total surface area and
volume.

6. (a) The given diagram is a solid prism of triangular base. 8 cm
If the volume of the prism is 480 cm³, find its height. 10 cm

Oasis School Mathematics-10 127

B E

(b) In the given figure, the volume of the triangular 13 cm D
4 cm 3 cm
prism is 18 cm³, AB = 4 cm and DF = 3 cm. Find the A

height of the prism. F

C

7. (a) The lateral surface area of the given triangular A A'
prism is 1800 cm2. Find the height of the prism. 5 cm
B B'

12 cm C'
C

(b) The base of the given triangular prism is an equilateral A 40 cm A'
triangle. If its lateral surface area is 1,200 cm², find the
B' C'
length of a side of the base.

B

C

8. (a) Total surface area of the given triangular prism is A 9 cm B A'
B'
1,188 cm². Find the height of the prism. 12 cm
C C'


(b) Total surface area of the given triangular prism 24 cm
is 3,528 cm². Find the height of the prism. 25 cm

9. (a) In the given figure, AB = 8 cm, AA' = 20cm, A 20cm A'

AC = 10cm and the area of rectangular surface of B 8cm B'
the prism is 480 cm2, find the length of BC. 10 cm

C C'

(b) The volume of given triangular prism is 270cm3, A 30cm A'

its length AA' = 30 cm and BC = 3 cm, find the B B'
length of AB and AC. C'
3cm
Answers
C

1. Consult your teacher 2. (a) 750 cm3 (b) 12 cm (c) 1,000 cm² (d) 18 cm (e) 440 cm²
(f) 180 cm² (g) 280 cm² 3. (a) 288 cm3 (b) 120 cm3 (c) 576 cm3 (d) 146.83 cm3 (e)1,800cm3
(f) 2,500 cm3 4. (a) 120 cm², 132 cm² (b) 900 cm², 960 cm² (c) 38.6 cm², 46.59 cm²
(d) 108 cm², 161.67 cm² (e) 180 cm², 211.17 cm² (f) 1,800 cm², 1,920 cm²

5. (a) 132 cm² (b) 445.7 cm², 360 cm3 6. (a) 20 cm (b) 3 cm 7. (a) 60 cm (b) 10 cm
8. (a) 30 cm (b) 60 cm 9. (a) 6 cm (b) 6cm, 6.7cm

128 Oasis School Mathematics-10

7.3 Pyramid

Pyramids of Egypt vertex

Look at the different types of pyramids. height
(i) If it has a triangular base, slant height
it is a triangular-based pyramid.
base
vertex

(ii) It has a rectangular base. So it is a slant height

rectangular-based pyramid. height

base

vertex

(iii) It has a square base. So it is called a square-based height
pyramid. It is also called regular pyramid. slant height

base

vertex

(iv) If a pyramid has a pentagon as a base, hesilgahntt

then it is called a pentagonal-based pyramid. height

base vertex

height

(v) If a pyramid has a hexagonal base, it is a Slant base
hexagonal-based pyramid. height

From the above figures, we can obtain the following conclusions.

• The base of the pyramid may be triangular, square, rectangular, polygonal, etc.
• Every lateral surface of a pyramid is triangular.
• Each triangular shape of the lateral surface has a common vertex.

• The height of a pyramid is perpendicular to the base.

Oasis School Mathematics-10 129

Volume of a pyramid

Prepare a net of a pyramid as shown in the figure.

Fold and paste them to form a pyramid. Make three
pyramids of the same size. Paste 3 Pyramids, then a cube is

formed.

Now,

3 × volume of the pyramid = Volume of a cube

Volume of the pyramid = 1 volume of the cube
3
1
= 3 area of base × height

= 1 a2h
3

Hence, the volume of the pyramid

= 1 area of base × height = 1 a2h
3 3

Lateral surface and total surface area of square-based pyramid

The given figure is a square-based pyramid, where ABCD is its base. OM is its slant

height. ON is its height. O

Let, AB = BC = CD = AD = a e

OM = l and ON = h l

Area of ∆OBC = 1 BC × OM Ah B
2 M
a2
= 1 × a × l.
2 N

Lateral surface area (LSA) = 4 × ∆OBC

=4× 1 al. D aC
2
BD = d
= 2al d
2
Total surface area (TSA) = LSA + area of base = 2al + a2 DN = NB =

∴ TSA of square-based pyramid = 2al + a2

Where a = length of a side of the base,

l = slant height of pyramid (height of the triangle)

In the given figure, ABM is a right angled triangle.

So, OB2 = OM2 + BM2 Note:

e2 = l2 + a­2 • If diagonal 'd' of
4 the base is

Similarly, in right angled triangle ONB, given, area of

OB2 = ON2 + NB2

e2 = h2 + d­2 base = 1 d2.
4 2

130 Oasis School Mathematics-10

Note:

• If 'e' be the length of the edge of the pyramid then,

e2 = l2 + ( a )2, e2 = l2 + a2
2 4

• If 'd' be the diagonal of the base, 'e' the length of the edge and h the vertical height of the

pyramid then,

e2 = h2 + ( d )2 = h2 + d2 .
2 4

Remember !

• Area of all triangular faces of a square-based pyramid is equal.

• Area of opposite triangular faces is equal in a rectangular pyramid.

In a square-based pyramid,

∆ OEF is a right-angled triangle. O

• l2 = h2 + a2 • e2 = l2 + a2 e
4 4
1 1 AB
• Area of ∆OBC = 2 BC × OF = 2 al EF
• EF = DC a2 Base
• Area 1 base = = 1 d2 if diagonal is given. DC
2 2
of

• LSA = 4 ∆OBC = 2al

• TSA = 4 ∆ + area of base = 2al + a2

• Volume = 1 area of ABCD × OE
3
1 1
= 3 area of base × height = 3 a2h

• Volume = 1 d2h, if diagonal 'd' is given.
6

Triangular-based pyramid

The given figure is a triangular-based pyramid having a base of an equilateral triangle.

In the given figure, the base of the pyramid is an equilateral triangle ABC. The sum of
the area of the triangular faces ∆OAB, ∆OBC and ∆OAC is the lateral surface area.

Lateral surface area = 3 ∆OAC O

= 3× 1 AC × OE A
2 DE
BC
Total surface area = LSA + area of base

= LSA + area of ∆ABC

OD is the height of the pyramid.

∴ Volume of the pyramid = 1 (area of base) × height
3

= 1 (area of ABC)×OD
3

Oasis School Mathematics-10 131

O
Worked Out Examples



Example: 1

Find the total surface area and volume of the given pyramid. A 6 cm D
Solution: P Q
In the given square-based pyramid,

OP is the height. ABCD is its base. B 16 cm C
Draw OQ ^ BC and join PQ.

Base area = 16 × 16 cm2 = 256 cm2 Alternative method
In ∆OPQ right-angled at P, Here, aside of the base (a) =16cm
Height (h) = 6 cm

OQ = OP² + PQ² We have, volume (v) = 1 a2h
3
1
= 6² + 8² [∵ PQ = 1 BC] = 3 ×162×6
2
= 100 = 10 cm. = 512 cm3

So, total surface area of the given pyramid = Area of the Again, l2 = h2 + a2
base + Area of 4 triangular faces 4
= (6)2 + (16)2

= 256 cm2 + 4 × ( 1 OQ × DC) = 36 + 64 4
2
= 100

= 256 cm2 + 4 × 1 × 10 × 16 cm2 ∴ l = 10 cm.
2
We have, TSA = a2 + 2al
= 256 cm2 + 320 cm2 = 576 cm2
= 162 + 2×10×16

Again, volume of the pyramid = 1 × area of = 256+576
3 = 576 cm2
1
base×height = 3 × 256 × 6 cm2 = 512 cm3

Example: 2 24 cm 25 cm

Find the lateral surface area and total surface area of the given O
square-based pyramid.

Solution:

Here O is the vertex and square ABCD is the base of the
pyramid. From O, draw OF perpendicular to the base and
OE⊥BC. Join EF.

Here, OE = 25 cm, OF = 24 cm A B
F E
Using the Pythagoras Theorem,
D C
FE = OE² – OF ²

= (25)² – (24)²

= 625 – 576

= 49 = 7 cm

132 Oasis School Mathematics-10

∴ DC = 2 FE = 2 × 7cm = 14 cm Alternative method

∴ DC = BC = 14 cm 1 Here, l = 25cm, h = 24 cm
2
Now, area of ∆OBC = BC × OE We have, l2 = 242 + a2
4
1
= 2 × 14 × 25 or, 252 = 2a2 + a2
4
a2
= 175 cm² or, 4 = 49

or, a2 = 49 × 4

We have, lateral surface area of pyramid (LSA) ∴ a = 14 cm

= 4∆OBC We have, LSA = 2al
= 4 × 175 cm² = 700 cm²
Again, Total surface area = LSA + area of base = 2×14×25cm2
= 700 cm² + (14)² cm²
= 700 cm2

TSA = 2al + a2

= 700 + (14)2

= (700 + 196) cm² = 700 + 196

= 896 cm² = 896 cm2

Example: 3

Find the volume of the given triangular-based pyramid. O

Solution: 129 3 cm

The given figure is a triangular-based pyramid having an DA

equilateral triangle on its base. 3 B C

Here, area of equilateral ∆ABC = 4 (6)² cm² 6 cm

3
= 4 × 36 cm²
= 9 3 cm2

Height of the pyramid = 129 3 cm

We have, Volume of the pyramid = 1 area of base × height
3

= 1 × 9 3 × 129 3 cm3
3

= 108 cm3

Example: 4

In the given square-based pyramid, DC = 18 cm, O
TSA = 864 cm². Find the height of the pyramid. AB
Solution: D 18 cm C

In the given square-based pyramid, DC = BC = 18 cm.

Draw OF perpendicular to the base and OE ^ BC. Join FE.

Area of base = (18)² = 324 cm²

Oasis School Mathematics-10 133

Total surface area (TSA) = 864 cm² O

We have, TSA = LSA + area of base

864 = 4 × ∆OBC + 324 A B
F E
or, 864 – 324 = 4 ∆OBC C
D
or, 540 = 4 ∆OBC
or, 540 = ∆OBC Alternative method
or, = ∆OBC
4

135

Area of ∆OBC = 135 cm² Here, a = 18 cm

or, 1 BC × OE = 135 We have, TSA = 2al + a2
2
or, 864 = 2 × 18l + (18)2
1
or, 2 ×18 × OE = 135 or, 864 = 36l + 324

or, 9 OE = 135 or, 36l = 864–324

or, OE = 15 cm or, l = 540
∴ 36

Again, using Pythagoras Theorem, l = 15 cm

OF = OE² – EF ² Again, l2 = h2 + a2
4
or, (15)2 = h2 + (18)2
= (15)² – (9)² 4

or, 225 = h2 + 81

= 225 – 81 or, h2 = 225–81

= 144 = 12 cm or, h2 = 144
∴ Height of the pyramid = 12 cm.
or h = 12 cm.

Example: 5

Slant height of a square-based pyramid is 13 cm and the total surface area is 360 cm².

Find the length of the side of the square base.

Solution: O

Let ABCD be the base of the pyramid. O its vertex, OE its slant height. A 13 cm

Let, each side of the square base = a B
E
Here, 1 D C
area of ∆OBC = 2 BC × OE

1
= 2 a × 13

LSA = 4 ∆OBC

134 Oasis School Mathematics-10

13 Alternative method
= 4 × 2 a cm² Here, l = 13 cm, TSA = 360 cm2
We have,
= 26 a cm² TSA = a2 + 2al
360 = a2 + 2a × 13
Again, TSA = LSA + area of base or, a2 + 26a – 360 = 0
or, a2 + (36–10)a–360 = 0
360 = 26 a + a² or, a2 + 36a–10a–360 = 0
or, a(a+36)–10(a+36) = 0
or, a² + 26a – 360 = 0 or, (a+36) (a-10) = 0
Either, a + 36 = 0
or, a² + (36 – 10) a – 360 = 0 a = -36 (not possible)
or, a–10 = 0 ⇒ a = 10cm.
or, a² + 36 a – 10 a – 360 = 0

or, a(a + 36) – 10(a + 36) = 0

or, (a + 36) (a – 10) = 0

Either, a + 36 = 0 ⇒ a = – 36 (not possible)

Or, a – 10 = 0 ⇒ a = 10 cm

∴ Length of each side of base = 10 cm

Exercise 7.3

1. (a) What is the formula of area of the the base of the square-based pyramid?
(b) What is the formula of the area of the base of a pyramid having an equilateral

triangle on its base?
(c) If the area of the base and height of the pyramid are given, write the formula to

find the volume of pyramid.
(d) If 'a' be a side of the base of a square-based pyramid and 'h' its height, write the

formula for the volume of the pyramid in terms of 'a' and 'h'.
(e) If 'a' be a side of the base of a pyramid whose base is an equilateral triangle and

h the height of the pyramid, write the formula for the volume of the pyramid in
terms of 'a' and 'h'.
(f) If 'A' represents the area of a triangular face of the pyramid, then what is the LSA
of (i) Square based pyramid (ii) Triangular-based pyramid
(g) If 'a' be a side of the base of square based pyramid and 'l' the slant height, write the
formula for the LSA and TSA of the pyramid.

2. (a) Find the volume of a square-based pyramid whose area of the base is 100 cm² and
the height is 12 cm.

(b) Find the volume of a square-based pyramid whose height is 10 cm and the area of
the base 60 cm².

(c) Find the volume of a triangular-based pyramid whose height is 15 cm and the

Oasis School Mathematics-10 135

area of the triangular base is 209 3 cm².

3. (a) Find the lateral surface area of a square-based pyramid in which the area of a
triangular face is 36 cm².

(b) Find the lateral surface area of a triangular-based pyramid in which the area of a
triangular face is 40cm2.

(c) Find the total surface area of a square-based pyramid whose area of the base is
100 cm² and area of a triangular face is 65 cm².

(d) Find the total surface area of a pyramid having an equilateral triangle on its base,
whose area of the base is 9 3 cm² and area of each triangular face is 15 cm².

4. Find the volume of the given pyramids.
(a) O (b) O (c) O

18 cm 10 cm P 13 cm Q
T U
AB PS 12 cm

D 10 cm C Q 15 cm R S R

(d) Q (e) (f)

10 cm
15 cm 8 cm 10 cm
12 cm EF

I 16 cm

HG 16 cm 12 cm

13 cm 25 cm O

(g) (h) i)24 cm A B
E

10 cm D C

5. Find the volume of the given triangular-based pyramids.

(a) O (b)

A9 3 cm 12 cm

6 cm

B 9 cm C

136 Oasis School Mathematics-10

6. Find the lateral surface area and the total surface area of the following pyramids.

(a) (b) (c) O

25 cm

15 cm

10 cm
DC

24 cm
APB
12 cm 12 cm
(f)
(e) O
(d) 17 cm


AD 15 cm
12 cm
6 cm O
16 cm

16 cm B 10 cm C

(g) (h) O

13 cm
25 cm
PQ
24 cm
T

S 10 cm R

7. Find the lateral surface area and total surface area of the given triangular pyramid.

(a) (b) O

12 cm P

10 cm QR
8 3 cm

8. (a) In the given figure, the total surface area of the given square 6 cm
based pyramid is 96 cm2 and a side of the square base is 6 cm.
Find the slant height of the pyramid.

(b) Total surface area of the given square-based pyramid 6 cm
is 384 cm². Find the slant height and height of the 8 cm
pyramid.

12 cm

(c) The given figure is a square-based pyramid. Total surface area
of the pyramid is 144 cm². If the length of each side of the square
base is 8 cm, find- (i) its slant height (ii) its height (iii) its volume.

Oasis School Mathematics-10 137

9. (a) In the given figure, the length of each side of the square base is
14 cm. If the volume of the pyramid is 1,568 cm3, find the

(i) height of the pyramid.

(ii) TSA of the pyramid. 14 cm


(b) In the given figure, the length of each side of the square base is O
10 cm. If its volume is 400 cm3, find

(i) its height. A B
(ii) its slant height.

(iii) its lateral surface area. D 10 cm C

10. (a) Total surface area of the given square-based pyramid is 96 cm². 5 cm
If its slant height is 5 cm, find its height.
15 cm

(b) Total surface area of the given square-based pyramid is 864 cm². O
If its slant is 15 cm, find

(i) length of each side of the base.
(ii) height of the pyramid.
(iii) volume of the pyramid.

11. (a) Volume of the given square-based pyramid is 1568 cm3. AB
If EF = 7 cm, calculate the area of the triangular surface. EF

D OC

(b) Volume of the given square based pyramid is 1,280 cm3. P Q
If MN = 8 cm. Calculate the TSA of the pyramid. M N

12. (a) In the given figure, the length of each side of the square S R
base is 8 cm. If the area of the triangular faces is 80 cm2, O
find the length of OE and the volume of the pyramid. A
ED
(b) The given figure is a square based pyramid, where
UT = 6cm, area of triangular faces is 240 cm2. Find the B 8 cm C S
length of OT, OU and volume of pyramid. O PU

QT R

138 Oasis School Mathematics-10

13. The vertical height and length of base of the square-based pyramid are in the ratio of
2 : 3. If total surface area of the pyramid is 384 sq.cm., find the slant height.

Answers

1. Consult your teacher 2. (a) 400 cm3 (b) 200 cm3 (c) 173.2 cm3 3. (a) 144 cm²
(b) 120 cm² (c) 360 cm² (d) 60.58 cm² 4. (a) 600 cm3 (b) 750 cm3 (c) 400 cm3
(d) 1296 cm3 (e) 512 cm3 (f) 253.92 cm3 (g) 363.62 cm3 (h) 784 cm3 (i) 256cm3

5. (a) 182.25 cm3 (b) 62.35 cm3 6. (a) 240 cm², 384 cm² (b) 360 cm², 504 cm² (c) 672 cm²,

868 cm² (d) 320 cm², 576 cm² (e) 260 cm², 360 cm² (f) 544 cm², 800 cm² (g) 700 cm², 896 cm²

(h) 240 cm², 340 cm² 7. (a) 180 cm², 223.3 cm² (b) 249.41 cm², 332.55cm² 8. (a) 5 cm (b) 10 cm, 8 cm

(c) (i) 5 cm, (ii) 3cm (iii) 64 cm3 9. (a) 24 cm, 896 cm² (b) 12 cm, 13 cm, 260 cm²

10. (a) 4 cm (b) 18 cm, 12 cm, 1296 cm3 11. (a) 700 cm2 (b) 800 cm2

12. (a) 5cm, 64cm3 (b) 10cm, 8cm, 384 cm3 13. 10 cm.

7.4 Combined Solids of Prism and Pyramid

The given figure is a combined solid of a cuboid (square-based prism) and a square-

based pyramid. O
As we know that the volume of a combined solid

= Volume of prism + Volume of pyramid h1 A D
C
= Area of base × height (h1) + 1 area of base × height (h2) B
3 a

While calculating the total surface area of the combined solid, h2 E

TSA = LSA of pyramid + LSA of prism + Area of base

LSA of combined solid = LSA of prism + LSA of pyramid. Ea G

Remember !

• LSA of pyramid = 4∆ = 2al

• LSA of cuboid (prism) = perimeter of base × height = 4ah
• LSA of combined solid = LSA of cuboid (prism) + LSA of pyramid
• TSA of combined solid = LSA of pyramid + LSA of cuboid (prism)+area of base.

• Volume of Prism = Area of base × height = a2h.

• Volume of Pyramid = 1 Area of base × height = 1 a2h.
3 3

• Volume of combined solid = Volume of Prism + Volume of Pyramid.

= a2h2 + 1 a2h1
3

Worked Out Examples

Example : 1

1. Find the volume of the given combined solid. 21cm
Solution:
12cm

This is the combined solid of a prism (cuboid) and pyramid.

10cm

Oasis School Mathematics-10 139

For prism Alternative method
For Prism,
Area of base = (10 cm)2 = 100 cm2 h = 12cm, a = 10 cm
Volume of Prism (V1) = a2h
height = 12 cm = 102 × 12
= 1200 cm3
We have, For Pyramid,

Volume (V1) = Area of base × height

= 100 × 12 = 1,200 cm3

For pyramid

Area of base = (10 cm)2 = 100 cm2 a = 10 cm, h = 21cm – 12cm = 9cm

Height = 21cm – 12 cm = 9 cm. Volume of Pyramid (V2) = 1 a2h
We have, 3
= 1 102 × 9
3
Volume of pyramid (V2) = 1 area of base × height = 300 cm2
3

= 1 × 100 × 9 Volume of combined
3
solid = V1 + V2
= 900 cm3 = 300cm3 = (1200 + 300) cm3
3

Hence, the volume of given solid (V) = V1 + V2 = 1500 cm3

= 1200 cm3 + 300 cm3 = 1500 cm3

Example: 2

Find the volume of the given combined solid.

Solution:

This figure is combined solid of two pyramids whose combined height is 30 cm.

Let, h1 = height of upper pyramid

h2 = height of lower pyramid

i.e. h1 + h2 = 30 cm 8cm
Area of base = (8)2 = 64 cm2
30cm

Volume of upper pyramid (V1) = 1 area of base × height
3

= 1 × 64 × h1
3

= 64h1 cm3 Alternative method
3
Area of base = (8 cm)2
Volume of lower pyramid (V2)

= 1 area of base × height = 64 cm2
3
Sum of height = 30 cm
1
= 3 × 64 × h2 We have,

= 64h2 cm3 Volume = 1 area of base ×
3 3
sum of the height
Volume of combined solid = V1 + V2
= 1 × 64 × 30 cm3
= 64h1 + 64h2 3
3 3
= 640 cm3

140 Oasis School Mathematics-10

= 64 (h1 + h2)
3

= 64 × 30
3

= 640 cm3 O
A 10cm
Example : 3 B
Find the lateral surface area and total surface area of the given solid.
I 16cm
D C

HG

Solution: E 12cm F

This is the combined solid of a square-based prism (cuboid) and a pyramid.

ABCD is the base of the pyramid and EFGH is the base of the prism. Both of them
are equal in area. OI is the slant height of the pyramid.

For the pyramid, BC = 12 cm, OI = 10 cm

Area of ∆OBC = 1 BC × OI Alternative method
2 For Prism,
Perimeter of the base (P) = 4a
Now, = 1 × 12 × 10 = 4 × 12 = 48 cm
2 Height (h) = 16 cm
We have,
= 60 cm2

LSA of the pyramid = 4 × 60 cm2 = 240 cm2

For prism (cuboid) LSA of prism (A1) = p × h
= 48 × 16cm2

Perimeter of the base = 4 × 12 cm = 48 cm. = 768 cm2

LSA of the prism = Perimeter of the base × height For Pyramid,
a = 12 cm, l = 10 cm.

= 48 × 16 cm2 = 768 cm2 We have,
Again,
LSA of pyramid (A2) = 2al
= 2 × 12 × 10

Area of base = (12)2 = 144 cm2­ = 240 cm2
Area of base (A3) = a2

Now, = 122

= 144 cm2

LSAof the given figure = LSAof pyramid + LSAof prism. Now, LSA of given figure

= 240 cm2 + 768 cm2 = A1 + A2
= 768 + 240 = 1008 cm2

= 1008 cm2 TSA of given figure

Again, = (1008+144)cm2
= 1,152 cm2.

TSA of the given figure = LSA of pyramid + LSA of
prism + Area of base

= (240 + 768 + 144) cm2 = 1152 cm2

Oasis School Mathematics-10 141

Example : 4

Find the lateral surface area and total surface area of the given combined solid.
Solution:

The given figure is the combined solid of a square-based prism (cuboid) and a
square based pyramid. ABCD is the base of the pyramid and EFGH the base of the
prism OM is the height of the pyramid. ON is its slant height.

For prism

Perimeter of the base = 4 × 6 cm = 24 cm O
Height of the prism = 10cm

Lateral surface area of the prism = p × h

= 24 cm × 10 cm = 240 cm2 14cm A B
For pyramid M C N 10cm
E

Again, height of the pyramid = (14 – 10) cm HG

= 4 cm 6cm

E 6cm F

∴ OM = 4 cm, MN = 1 DC
2

= 1 × 6 cm = 3cm.
2

Using the Pythagoras Theorem in ∆OMN, Alternative method

ON2 = OM2 + MN2 For the pyramid,

= (4)2 + (3)2 l2 = h2 + a2
or, 4
= 16 + 9 = 25.
l2 = 42 + 62
∴ ON = 5 cm. 4

1 or, l2 = 25
2
Area of ∆OBC = BC × ON or, l = 5 cm

= 1 × 6×5 cm2 LSA of the Pyramid = 2al
2 = 2× 6 × 5
= 60 cm2
= 15 cm2 LSA of the Prism = 4ah
= 4 × 6 × 10cm2
∴ LSA of pyramid = 4∆ = 240 cm2

= 4 × 15 cm2 = 60 cm2.

∴ LSA of given solid LSA of the given figure
= 60cm2+240cm2
= LSA of prism + LSA of pyramid = 300 cm2

= 240 cm2 + 60 cm2 = 300 cm2 Area of base = (6)2cm2 = 36cm2

TSA of the given solid = LSA + area of base TSA of the given figure

= 300 + (6)2 = 300cm2 + 36cm2

= 300 + 36 = 336cm2

= 336 cm2

142 Oasis School Mathematics-10

Example: 5

The given figure is the combined solid of square based prism and 13cm
pyramid, where the slant height of the pyramid is 13 cm and the length
of the side of the base is 10cm. If the volume of whole solid is 900 cm3, 10cm
find the height of the prism.

Solution:

Given figure is the combined solid of square based prism and pyramid.

Here, to find the volume of prism,

Let the height of the prism be h1
Then, area of the base of the prism = (10 cm)2

= 100 cm2

We have,

Volume of the prism (V1) = Area of base × height
= 100 × h1
= 100h1 cm3.
Again, for pyramid,

Slant height (l) = 13 cm.

Length of a side of the base (a) = 10 cm.

We have, l2 = h2 + a2
4

or, 132 = h2 + 102
4

or, 169 = h2 + 25

or, 169 – 25 = h2

or, h2 = 144

∴ h = 144 = 12 cm.

Again, we have

Volume of the pyramid (V2) = 1 a2h
= 3
1
3 102 × 12

= 400 cm3.

Now,

Volume of whole solid (V) = V1 + V2

or, 900 = 100h1 + 400

or, 900 – 400 = 100h1

or, 500 = 100h1

or, h1 = 500 = 5 cm.
∴ Height of the prism = 100

5 cm.

Oasis School Mathematics-10 143

Exercise 7.3

1. Find the volume of the given combined solids.

(a) (b) (c) (d)

6cm 12cm
13cm
5cm 5cm 5cm
15cm
15cm

8cm

6cm 6cm 8cm 8cm
6cm
(e) (f) O O

20cm (g)

20cm A 18cm
C
A B B
10cm C
D A' 20cm

16cm 6cm 8cm

B' C'

2. The given figure is a combined solid of a square-based prism and a pyramid. From the

given data, find -

(i) Height of the pyramid. O

(ii) Slant height of the pyramid. HG

(iii) Area of a triangular face of the pyramid. 44cm E I 20cm
F

(iv) LSA of the pyramid. HC

(v) Perimeter of the base of the prism. A 14cm B
(vi) LSA of the prism.

(vii) Area of base of the prism.

(viii) LSA of the whole solid

(ix) TSA of the whole solid.

3. Find the lateral surface area and total surface area of the given combined solids.

(a) (b) (c) (d)
10cm
13cm 24cm 8cm

15cm

30cm 12cm 10cm 20cm
10cm
10cm 14cm

144 Oasis School Mathematics-10